Printable · GCSE Higher · ages 14-16

Algebra worksheet — GCSE Higher

Fifteen questions across the algebra statements at Higher tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.

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Algebra worksheet — GCSE Higher

Maths · GCSE Higher · Algebra · 15 questions

MathsUKwww.geekhero.co.uk

1.Priya has a budget of £50 for a school trip. The coach costs £14 and each student ticket costs £4. Using the inequality 14 + 4s ≤ 50, work out the greatest number of student tickets, s, she can buy.
(a)16
(b)9
(c)8
(d)12
2.Find the common ratio of the geometric sequence 5, 5√2, 10, 10√2, ... Give your answer in surd form.
(a)1.41
(b)√2
(c)5√2
(d)2
3.The first term of an arithmetic sequence is 5, and each term after that increases by 6. Work out the 8th term of the sequence.
(a)53
(b)41
(c)48
(d)47
4.The curve y = x² − 6 and the line y = 2x − 3 intersect at two points. Which pair of points is correct?
y = 2x − 3y = x² − 6
(a)(3, 9) and (−1, 1)
(b)(3, 3) and (−1, −5)
(c)(3, 3)
(d)(3, 3) and (1, −1)
5.The iterative formula xₙ₊₁ = 12 ÷ (xₙ + 2) is used repeatedly, starting from x₀ = 1. Work out the value that xₙ approaches, correct to 2 decimal places.
(a)3.00
(b)6.00
(c)2.61
(d)4.00
6.Work out the value of 5m² + 3m when m = −2.4
(a)36
(b)29.4
(c)21.6
(d)−36
7.Solve the inequality 6.3x + 1.8 ≤ 40. Hence write down the greatest number with exactly one decimal place that satisfies the inequality.
(a)6.6
(b)6.1
(c)6.3
(d)6.0
8.A circle has centre (0, 0) and equation x² + y² = 50. Work out the radius of the circle, giving your answer as a surd in its simplest form.
(a)50
(b)25
(c)25√2
(d)5√2
9.Ben is asked to find the inverse of f(x) = 4 − 3x. He writes f⁻¹(x) = (4 − x)/3. Which statement about Ben's answer is correct?
(a)Wrong: sign kept, giving f⁻¹(x) = (−4 − x)/3
(b)Correct, but only because f is its own inverse
(c)Correct: 3y = 4 − x gives f⁻¹(x) = (4 − x)/3
(d)Wrong: correct inverse is f⁻¹(x) = (x − 4)/3
10.Solve the simultaneous equations 3x + y = 14 and x + y = 6. Work out the value of x.
(a)4.67
(b)−4
(c)2
(d)4
11.The solution set of an inequality is described as: x is less than −3, or x is greater than or equal to 1. Write this solution set using set notation.
(a){x : x < −3} ∪ {x : x ≥ 1}
(b){x : −3 < x ≤ 1}
(c){x : x > −3} ∪ {x : x ≤ 1}
(d){x : x ≤ −3} ∪ {x : x > 1}
12.A proof that the product of two consecutive even numbers is always a multiple of 8 begins: Let the two consecutive even numbers be 2n and 2n + 2, so their product is 2n(2n + 2) = 4n(n + 1). Which line correctly completes the proof?
(a)4n(n + 1) is a multiple of 4, and because n and n + 1 are consecutive integers, it must be a multiple of 8.
(b)Expanding gives 4n(n + 1) = 4n² + 4n, which is clearly a multiple of 8.
(c)4n is always a multiple of 4 for any integer n, so 4n(n + 1) is a multiple of 4 as well, which means it is a multiple of 8.
(d)n and n + 1 are consecutive integers, so one of them must be even; this makes n(n + 1) even, so 4n(n + 1) is 4 × an even number, which is a multiple of 8.
13.Solve (3x + 2)/5 = (x + 4)/3
(a)x = 3.5
(b)x = 4.5
(c)x = 10
(d)x = 1
14.The diagram shows a distance–time graph for a cyclist travelling at a constant speed, for the first 6 minutes of a journey. Distance is in kilometres and time is in minutes. Work out how far the cyclist would travel in 20 minutes at the same speed.
(a)20 km
(b)10 km
(c)40 km
(d)3 km
15.A car's journey is represented by a velocity-time graph. The total distance travelled, found from the area under the graph, is 350 metres, and the total time for the journey is 25 seconds. Work out the car's average speed for the whole journey, in m/s.
(a)375 m/s
(b)14 m/s
(c)8750 m/s
(d)0.07 m/s

Answer key

(b) 9 — Subtract 14 from both sides: 4s ≤ 36. Divide both sides by 4: s ≤ 9, so the greatest number of tickets is 9. A candidate who forgets the £14 coach cost solves 4s ≤ 50, getting s ≤ 12.5, rounded down to 12. A candidate who adds the £14 instead of subtracting it solves 4s ≤ 64, getting s = 16. A candidate who miscalculates 50 − 14 as 32 solves 4s ≤ 32, getting s = 8.
(b) √2 — 5√2 ÷ 5 = √2. Checking between the third and second terms: 10 ÷ 5√2 = √2 as well (since 10 ÷ 5√2 = 2 ÷ √2 = √2), so the common ratio is confirmed as √2 throughout. Squaring the ratio instead of leaving it as a surd gives 2, which is wrong because 2 is the SQUARE of the common ratio, not the ratio itself. Rounding the exact surd to a decimal gives 1.41, which is wrong because the sequence is defined using an exact surd ratio, and a rounded decimal is not the same value. Writing down the SECOND TERM of the sequence, 5√2, instead of the ratio BETWEEN terms, gives 5√2, which is wrong because a term of the sequence is not the same thing as the common ratio.
(d) 47 — To reach the 8th term from the 1st term, the difference of 6 is added 7 times (8 − 1 = 7): 5 + 7 × 6 = 47. A candidate who multiplies by the term number itself, rather than one less, would compute 5 + 8 × 6 = 53. A candidate who uses one step too few (6 instead of 7) would reach 5 + 6 × 6 = 41. A candidate who forgets to include the first term at all would compute just 8 × 6 = 48.
(b) (3, 3) and (−1, −5) — Set the two expressions for y equal: x² − 6 = 2x − 3, which rearranges to x² − 2x − 3 = 0. Factorise: (x − 3)(x + 1) = 0, so x = 3 or x = −1. Substitute into the linear equation, y = 2x − 3: x = 3 gives y = 3; x = −1 gives y = −5. The points are (3, 3) and (−1, −5). Distractor routes: (3, 3) and (1, −1) comes from mis-factorising x² − 2x − 3 as (x − 3)(x − 1), giving a second root of 1 instead of −1. (3, 9) and (−1, 1) comes from finding the y-coordinate from y = x² instead of substituting back into the given line equation y = 2x − 3. (3, 3) alone stops after finding only the first root of the quadratic.
(c) 2.61 — The limit L satisfies L = 12 ÷ (L + 2), so L(L + 2) = 12, giving L² + 2L − 12 = 0 and L = (−2 + √52) ÷ 2 = 2.6056, which is 2.61 to 2 decimal places (the early iterates 4, 2, 3, 2.4, 2.73, ... oscillate around this value before settling). Choosing 3.00 reads off x₃, one of the early oscillating values, before the sequence has settled close to the limit. Choosing 4.00 reads off x₁, the very first iterate, not the value the sequence approaches. Choosing 6.00 comes from writing the limit equation as L = 12 ÷ 2, leaving L itself out of the denominator.
(c) 21.6 — m² = (−2.4)² = 5.76, since a negative number squared is positive, so 5m² = 28.8. Then 3m = 3 × (−2.4) = −7.2. Total: 28.8 + (−7.2) = 21.6. −36 comes from taking m² as negative, 5 × (−5.76) − 7.2 = −36, forgetting that squaring removes the negative sign. 36 comes from ignoring the negative sign on m altogether and using m = 2.4 throughout. 29.4 comes from treating 3m as 3 + m instead of 3 × m, giving 28.8 + (3 + (−2.4)).
(d) 6.0 — Subtract 1.8 from both sides: 6.3x ≤ 38.2. Divide both sides by 6.3: x ≤ 6.0634… . The question asks for the greatest value to 1 decimal place that still satisfies the inequality. Testing 6.1: 6.3 × 6.1 + 1.8 = 40.23, which is more than 40, so 6.1 fails. Testing 6.0: 6.3 × 6.0 + 1.8 = 39.6, which is no more than 40, so 6.0 works and is the greatest such value. A candidate who simply rounds 6.0634… to 1 decimal place answers 6.1, without checking that it satisfies the inequality. A candidate who adds 1.8 instead of subtracting works out 41.8 ÷ 6.3 and answers 6.6. A candidate who forgets the 1.8 altogether divides 40 by 6.3 and answers 6.3.
(d) 5√2 — For a circle x² + y² = r², the 50 on the right-hand side is r², not r, so the radius is √50. Writing 50 as 25 × 2, the largest square factor times what remains, gives √50 = √25 × √2 = 5√2. Forgetting to square-root 50 at all and giving the value of r² instead gives 50. Halving 50 instead of taking its square root gives 25. Using 25 as the number left outside the square root sign, instead of as the number under it, gives the wrongly simplified 25√2.
(c) Correct: 3y = 4 − x gives f⁻¹(x) = (4 − x)/3 — Swap x and y: x = 4 − 3y. Add 3y to both sides and subtract x from both sides: 3y = 4 − x. Divide by 3: y = (4 − x)/3, which is exactly what Ben wrote — his rearrangement is correct. Check with a value: f(1) = 4 − 3 = 1, and Ben's formula gives (4 − 1)/3 = 1, which matches. 'Correct, but only because f is its own inverse' gives the right verdict for a false reason — f(f(x)) = 4 − 3(4 − 3x) = 9x − 8, which is not x, so f is not self-inverse; Ben's rearrangement is correct for the ordinary algebraic reason above, not because of any special property of f. 'Wrong: sign kept, giving (−4 − x)/3' comes from not carrying the swap through consistently — testing x = 1 gives (−4 − 1)/3 = −5/3, which does not equal 1, so it is wrong. 'Wrong: correct inverse is (x − 4)/3' comes from writing 3y = x − 4 instead of 3y = 4 − x, a sign slip when isolating y — testing x = 1 gives (1 − 4)/3 = −1, which again does not equal 1.
(d) 4 — Subtracting the second equation from the first: the y-terms cancel, and the x-terms combine as 3x − x = 2x; the right-hand sides give 14 − 6 = 8. This gives 2x = 8, so x = 4. A candidate who subtracts in the wrong order would get 2x = 6 − 14 = −8, so x = −4. A candidate who uses only the first equation, dividing 14 by 3 as if y were 0, would get x ≈ 4.67. A candidate who reports the value of y instead of x would get y = 6 − 4 = 2.
(a) {x : x < −3} ∪ {x : x ≥ 1} — "Less than −3" stays strict, since the wording never says "or equal to": x < −3. "Greater than or equal to 1" is inclusive: x ≥ 1. These are two separate, non-overlapping ranges joined with "or", so in set notation they are combined with the union symbol: {x : x < −3} ∪ {x : x ≥ 1}. Distractor routes: {x : x ≤ −3} ∪ {x : x > 1} swaps the strict and inclusive signs, marking −3 as included and 1 as excluded, the opposite of the wording. {x : −3 < x ≤ 1} treats "or" as "and", joining the two conditions into one continuous interval between the values instead of a union of two separate ranges. {x : x > −3} ∪ {x : x ≤ 1} reverses both inequality directions; the two reversed ranges then overlap and between them cover every number on the number line, so that set is the whole of the real line rather than the two separate ranges the description asks for.
(d) n and n + 1 are consecutive integers, so one of them must be even; this makes n(n + 1) even, so 4n(n + 1) is 4 × an even number, which is a multiple of 8. — The proof needs a reason why n(n + 1) is even, not just an assertion. n and n + 1 are consecutive integers, so exactly one of them is even; multiplying by that even number keeps n(n + 1) even, so 4n(n + 1) = 4 × (an even number), and 4 × an even number always has a further factor of 2 hidden inside it, making the whole product a multiple of 8. The option 'is a multiple of 4, and because n and n + 1 are consecutive integers, it must be a multiple of 8' asserts the multiple-of-8 conclusion directly from 'multiple of 4' and 'consecutive integers' without ever showing that n(n + 1) itself is even — the missing step is exactly what earns the mark. The option '4n is always a multiple of 4 ... which means it is a multiple of 8' mistakes 4n being a multiple of 4 for the whole product 4n(n + 1) being a multiple of 8; that extra factor of 2 only comes from n(n + 1) being even, not from 4n alone. The option that expands to 4n² + 4n and calls it 'clearly a multiple of 8' never checks for a factor of 2 beyond the 4 already there — the word 'clearly' is standing in for a missing argument.
(a) x = 3.5 — Method: clear the fractions by multiplying both sides by 15, expand both brackets, then collect the x terms on one side and the numbers on the other. Working: multiplying both sides by 15 gives 3(3x + 2) = 5(x + 4), which expands to 9x + 6 = 5x + 20; subtracting 5x and 6 from both sides gives 4x = 14, and dividing both sides by 4 gives x = 3.5. Answer: x = 3.5. The distractors: x = 1 comes from collecting the x terms by adding the 5x instead of subtracting it, giving 9x + 5x = 20 − 6 and so 14x = 14; x = 4.5 comes from expanding 3(3x + 2) as 9x + 2, multiplying only the x term by the 3, which leads to 4x = 18; x = 10 comes from reaching 4x = 14 correctly and then subtracting 4 instead of dividing by 4.
(b) 10 km — Method: find the constant speed from the graph (distance ÷ time for any point on the line), then multiply that speed by 20 minutes. Working: the line passes through (4 minutes, 2 km), so the speed is 2 ÷ 4 = 0.5 km per minute; in 20 minutes the cyclist travels 0.5 × 20 = 10 km. Answer: 10 km. Distractor refutation: 3 km comes from reading off the distance shown at the end of the plotted section (6 minutes) and stopping there, instead of extending the line to 20 minutes. 20 km comes from misreading the speed as 1 km per minute instead of 0.5 km per minute, doubling the true rate. 40 km comes from dividing 20 by the speed instead of multiplying by it, a reciprocal mix-up.
(b) 14 m/s — Average speed for a whole journey is total distance divided by total time: 350 ÷ 25 = 14 m/s. Multiplying the distance and time instead of dividing gives 350 × 25 = 8750 m/s. Adding the distance and time instead of dividing gives 350 + 25 = 375 m/s. Inverting the division, working out time divided by distance, gives 25 ÷ 350, which rounds to 0.07 m/s.

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